1. Calorimetry

2. Energy released to the surrounding as heat
Burning of a Match
System
Surroundings
D(PE)
(Reactants)
Potential energy
Energy released to the surrounding as heat
(Products)
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 293

3. Conservation of Energy in a Chemical Reaction
In this example, the energy of the reactants and products increases,
while the energy of the surroundings decreases.
In every case, however, the total energy does not change.
Endothermic Reaction
Reactant + Energy Product
Surroundings
Surroundings
System
Energy
System
Before
reaction
After
reaction
Myers, Oldham, Tocci, Chemistry, 2004, page 41

4. Conservation of Energy in a Chemical Reaction
In this example, the energy of the reactants and products decreases,
while the energy of the surroundings increases.
In every case, however, the total energy does not change.
Exothermic Reaction
Reactant Product + Energy
Surroundings
Surroundings
System
System
Energy
Before
reaction
After
reaction
Myers, Oldham, Tocci, Chemistry, 2004, page 41

5. Direction of Heat Flow Surroundings ENDOthermic EXOthermic qsys > 0
System
ENDOthermic
qsys > 0
EXOthermic
qsys < 0
System
H2O(s) + heat  H2O(l)
melting
H2O(l)  H2O(s) + heat
freezing
Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 207

6. Caloric Values Food joules/grams calories/gram Calories/gram
Protein
Fat
Carbohydrates
1calories = joules
1000 calories = 1 Calorie
"science" "food"
Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51

7. Experimental Determination of Specific Heat of a Metal
Typical apparatus used in this activity include a boiler (such as large glass beaker), a heat source (Bunsen burner or hot plate), a stand or tripod for
the boiler, a calorimeter, thermometers, samples (typically samples of copper, aluminum, zinc, tin, or lead), tongs (or forceps or string) to handle samples, and a balance.

8. A Coffee Cup Calorimeter
Thermometer
Styrofoam
cover
cups
Stirrer
A Coffee Cup Calorimeter
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302

9. Bomb Calorimeter thermometer stirrer full of water ignition wire
steel “bomb”
sample

10. oxygen supply stirrer thermometer magnifying eyepiece air space
crucible
steel bomb
sample
ignition coil
bucket
heater
water
ignition
wires
insulating
jacket
1997 Encyclopedia Britanica, Inc.

11. A Bomb Calorimeter

12. Causes of Change - Calorimetry
Outline
Keys

13. Heating Curves Gas - KE  Boiling - PE  Liquid - KE  Melting - PE 
Solid - KE 
Courtesy Christy Johannesson

14. Heating Curves Gas - KE  Boiling - PE  Liquid - KE  Melting - PE 
140
Gas - KE 
120
100
Boiling - PE  80 60 40
Liquid - KE 
Temperature (oC) 20
Melting - PE 
-20
-40
Solid - KE 
-60
-80
-100
Time

15. Heating Curves Gas - KE  Boiling - PE  Liquid - KE  Melting - PE 
140
Gas - KE 
120
100
Boiling - PE  80 60 40
Liquid - KE 
Temperature (oC) 20
Melting - PE 
-20
-40
Solid - KE 
-60
-80
-100
Time

16. Heating Curves Temperature Change Heat Capacity
change in KE (molecular motion)
depends on heat capacity
Heat Capacity
energy required to raise the temp of 1 gram of a substance by 1°C
“Volcano” clip -
water has a very high heat capacity
Courtesy Christy Johannesson

17. Heating Curves Phase Change Heat of Fusion (Hfus)
change in PE (molecular arrangement)
temp remains constant
Heat of Fusion (Hfus)
energy required to melt 1 gram of a substance at its m.p.
Courtesy Christy Johannesson

18. Heating Curves Heat of Vaporization (Hvap)
energy required to boil 1 gram of a substance at its b.p.
usually larger than Hfus…why?
EX: sweating, steam burns, the drinking bird
Courtesy Christy Johannesson

19. Phase Diagrams
Show the phases of a substance at different temps and pressures.
Courtesy Christy Johannesson

20. Humor
A small piece of ice which lived in a test tube fell in love with a Bunsen burner.
“Bunsen! My flame! I melt whenever I see you” said the ice.
The Bunsen burner replied” “It’s just a phase you’re going through”.

21. Heating Curve for Water (Phase Diagram)
140
120
100 80 60 40 20
-20
-40
-60
-80
-100 F
Heat = m x Cvap
Cv = J/g D E BP
Heat = m x Cfus
Cf = 333 J/g
Heat = m x DT x Cp, gas
Cp (steam) = J/goC
Heat = m x DT x Cp, liquid
Temperature (oC)
Cp = J/goC MP B C
Heat = m x DT x Cp, solid
A  B warm ice
B  C melt ice (solid  liquid)
C  D warm water
D  E boil water (liquid  gas)
E  D condense steam (gas  liquid)
E  F superheat steam
Cp (ice) = J/goC A
Heat

22. Calculating Energy Changes - Heating Curve for Water
140
DH = mol x DHvap
120
DH = mol x DHfus
100 80 60
Heat = mass x Dt x Cp, gas 40
Temperature (oC) 20
Heat = mass x Dt x Cp, liquid
-20
-40
-60
Heat = mass x Dt x Cp, solid
-80
-100
Time

23. Equal Masses of Hot and Cold Water
Thin metal wall
Insulated box
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

24. Water Molecules in Hot and Cold Water
Hot water Cold Water
90 oC oC
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

25. Water Molecules in the same temperature water
(50 oC)
Water
(50 oC)
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

26. Heat Transfer Surroundings Final Temperature Block “A” Block “B”
SYSTEM
20 g (40oC)
20 g (20oC)
30oC Al Al
m = 20 g
T = 40oC
m = 20 g
T = 20oC
What will be the final temperature
of the system ?
a) 60oC b) 30oC c) 20oC d) ?
Assume NO heat energy is “lost” to the surroundings from the system.

27. Heat Transfer ? Surroundings Final Temperature Block “A” Block “B”
SYSTEM
20 g (40oC)
20 g (20oC)
30.0oC
20 g (40oC)
10 g (20oC)
33.3oC Al Al
m = 20 g
T = 40oC
m = 10 g
T = 20oC
What will be the final temperature
of the system ?
a) 60oC b) 30oC c) 20oC d) ?
Assume NO heat energy is “lost” to the surroundings from the system.

28. Heat Transfer Surroundings Final Temperature Block “A” Block “B”
SYSTEM
20 g (40oC)
20 g (20oC)
30.0oC
20 g (40oC)
10 g (20oC)
33.3oC
20 g (20oC)
10 g (40oC)
26.7oC Al Al
m = 20 g
T = 20oC
m = 10 g
T = 40oC
Assume NO heat energy is “lost” to the surroundings from the system.

29. Heat Transfer Surroundings Final Temperature Block “A” Block “B”
SYSTEM
20 g (40oC)
20 g (20oC)
30.0oC
20 g (40oC)
10 g (20oC)
33.3oC
20 g (20oC)
10 g (40oC)
26.7oC
H2O Ag
m = 75 g
T = 25oC
m = 30 g
T = 100oC
Real Final Temperature = 26.7oC
Why?
We’ve been assuming ALL materials
transfer heat equally well.

30. Specific Heat Water and silver do not transfer heat equally well.
Water has a specific heat Cp = J/goC
Silver has a specific heat Cp = J/goC
What does that mean?
It requires Joules of energy to heat 1 gram of water 1oC
and only Joules of energy to heat 1 gram of silver 1oC.
Law of Conservation of Energy…
In our situation (silver is “hot” and water is “cold”)…
this means water heats up slowly and requires a lot of energy
whereas silver will cool off quickly and not release much energy.
Lets look at the math!

31. Specific Heat
The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.

32. Calculations involving Specific HeatOR
cp = Specific Heat
q = Heat lost or gained
T = Temperature change
m = Mass

33. Table of Specific Heats
Specific Heats of Some Common
Substances at K
Substance Specific heat J/(g.K)
Water (l)
Water (s)
Water (g)
Ammonia (g)
Benzene (l)
Ethanol (l)
Ethanol (g)
Aluminum (s)
Calcium (s)
Carbon, graphite (s)
Copper (s)
Gold (s)
Iron (s)
Mercury (l)
Lead (s)

34. Latent Heat of Phase Change
Molar Heat of Fusion
The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point.
The energy that must be removed in order to convert one mole of liquid to solid at its freezing point.

35. Latent Heat of Phase Change #2
Molar Heat of Vaporization
The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point.
The energy that must be removed in order to convert one mole of gas to liquid at its condensation point.

36. Latent Heat – Sample Problem
Problem: The molar heat of fusion of water is kJ/mol. How much energy is needed to convert 60 grams of ice at 0C to liquid water at 0C?
Mass
of ice
Molar
Mass of
water
Heat of
fusion

37. Heat of Reaction
The amount of heat released or absorbed during a chemical reaction.
Endothermic:
Reactions in which energy is absorbed as the reaction proceeds.
Exothermic:
Reactions in which energy is released as the reaction proceeds.

38. Calorimetry “loses” heat Surroundings SYSTEM Tfinal = 26.7oC H2O Ag
m = 75 g
T = 25oC
m = 30 g
T = 100oC

39. Calorimetry Surroundings SYSTEM H2O Ag m = 75 g T = 25oC m = 30 g

40. 1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 BTU (British Thermal Unit) – amount of heat needed to raise 1 pound of water 1oF.
1 calorie - amount of heat needed to raise 1 gram of water 1oC
1 Calorie = calories
“food” = “science”
Candy bar
300 Calories = 300,000 calories
English
Metric = _______
Joules
1 calorie = Joules

41. q = Cp . m . DT Cp(ice) = 2.077 J/g oC
Temperature (oC) 40 20
-20
-40
-60
-80
-100
120
100 80 60
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp, liquid
Heat = mass x Dt x Cp, gas
Heat = mass x Dt x Cp, solid
Cp(ice) = J/g oC
It takes Joules to raise 1 gram ice 1oC.
X Joules to raise 10 gram ice 1oC.
(10 g)(2.077 J/g oC) = Joules
X Joules to raise 10 gram ice 10oC.
(10oC)(10 g)(2.077 J/g oC) = Joules
q = Cp . m . DT
Heat = (specific heat) (mass) (change in temperature)

42. Temperature (oC) 40 20
-20
-40
-60
-80
-100
120
100 80 60
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp, liquid
Heat = mass x Dt x Cp, gas
Heat = mass x Dt x Cp, solid
q = Cp . m . DT
Heat = (specific heat) (mass) (change in temperature)
Given Ti = -30oC
Tf = -20oC
q = Joules

43. Find the mass of the iron.
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC).
When thermal equilibrium is reached, the system has a temperature of 42oC.
Find the mass of the iron. Fe
T = 500oC
mass = ? grams
T = 20oC
mass = 240 g -
LOSE heat = GAIN heat
- [(Cp,Fe) (mass) (DT)] = (Cp,H2O) (mass) (DT)
- [( J/goC) (X g) (42oC - 500oC)] = (4.184 J/goC) (240 g) (42oC - 20oC)]
Drop Units:
- [(0.4495) (X) (-458)] = (4.184) (240 g) (22)
205.9 X =
X = g Fe
Calorimetry Problems 2
question #5

44. A 97 g sample of gold at 785oC is dropped into 323 g of water, which has an initial
temperature of 15oC. If gold has a specific heat of J/goC, what is the final
temperature of the mixture? Assume that the gold experiences no change in state
of matter. Au
T = 785oC
mass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat -
- [(Cp,Au) (mass) (DT)] = (Cp,H2O) (mass) (DT)
Drop Units:
- [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)]
- [(12.5) (Tf - 785oC)] = (1.35 x 103) (Tf - 15oC)]
-12.5 Tf x = x 103 Tf x 104
3 x 104 = x 103 Tf
Tf = oC
Calorimetry Problems 2
question #8

45. If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature
of the system.
T = 13oC
mass = 59 g
T = 72oC
mass = 87 g
LOSE heat = GAIN heat -
- [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT)
Drop Units:
- [(4.184 J/goC) (87 g) (Tf - 72oC)] = (4.184 J/goC) (59 g) (Tf - 13oC)
- [(364.0) (Tf - 72oC)] = (246.8) (Tf - 13oC)
-364 Tf = Tf
= Tf
Tf = oC
Calorimetry Problems 2
question #9

46. A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC.
Find the system's final temperature.
Temperature (oC) 40 20
-20
-40
-60
-80
-100
120
100 80 60
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp, liquid
Heat = mass x Dt x Cp, gas
Heat = mass x Dt x Cp, solid
ice
T = -11oC
mass = 38 g D
water cools
T = 56oC
mass = 214 g B
warm water A C
warm ice
melt ice
LOSE heat = GAIN heat - D A B C
- [(Cp,H2O(l)) (mass) (DT)] = (Cp,H2O(s)) (mass) (DT) + (Cf) (mass) + (Cp,H2O(l)) (mass) (DT)
- [(4.184 J/goC)(214 g)(Tf - 56oC)] = (2.077 J/goC)(38 g)(11oC) + (333 J/g)(38 g) + (4.184 J/goC)(38 g)(Tf - 0oC)
- [(895) (Tf - 56oC)] = (159) (Tf)]
- 895 Tf = Tf
- 895 Tf = Tf
= Tf
Tf = oC
Calorimetry Problems 2
question #10

47. qD = (4.184 J/goC) (238.4 g) (Tf - 8oC)
(1000 g = 1 kg)
25 g of 116oC steam are bubbled into kg of water at 8oC. Find the final temperature of the system.
238.4 g
- [qA + qB + qC] = qD
- [(Cp,H2O) (mass) (DT)] (-Cv,H2O) (mass) + (Cp,H2O) (mass) (DT) = [(Cp,H2O) (mass) (DT)]
qD = (4.184 J/goC) (238.4 g) (Tf - 8oC)
qD = 997Tf
qA = [(Cp,H2O) (mass) (DT)]
qB = (Cv,H2O) (mass)
qC = [(Cp,H2O) (mass) (DT)]
qA = [(1.87 J/goC) (25 g) (100o - 116oC)]
qB = (-2256 J/g) (25 g)
qC = [(4.184 J/goC) (25 g) (Tf - 100oC)]
qA = J
qB = J
qC = 104.5Tf
- [qA + qB + qC] = qD
- [ Tf ] = 997Tf
Temperature (oC) 40 20
-20
-40
-60
-80
-100
120
100 80 60
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp, liquid
Heat = mass x Dt x Cp, gas
Heat = mass x Dt x Cp, solid
Tf = 997Tf
Tf = 997Tf A
= Tf C
1102 B
Tf = 68.6oC D
Calorimetry Problems 2
question #11

48. (1000 g = 1 kg)
25 g of 116oC steam are bubbled into kg of water at 8oC. Find the final temperature of the system.
238.4 g
- [qA + qB + qC] = qD
- [(Cp,H2O) (mass) (DT) (-Cv,H2O) (mass) + (Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT)
- [(Cp,H2O) (mass) (DT) +
(Cv,H2O) (mass) +
(Cp,H2O) (mass) (DT)]
= (4.184 J/goC) (238.4 g) (Tf - 8oC)
- [(1.87 J/goC) (25 g) (100o - 116oC) +
(-2256 J/g) (25 g) +
(4.184 J/goC) (25 g) (Tf - 100oC)]
= 997Tf
- [ J J
Tf ]
= 997Tf
- [ Tf ] = 997Tf
Tf = 997Tf
Tf = 997Tf
Temperature (oC) 40 20
-20
-40
-60
-80
-100
120
100 80 60
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp, liquid
Heat = mass x Dt x Cp, gas
Heat = mass x Dt x Cp, solid
= Tf
1102
Tf = 68.6oC A C B D
Calorimetry Problems 2
question #11

49. A 322 g sample of lead (specific heat = 0
A 322 g sample of lead (specific heat = J/goC) is placed into 264 g of water at 25oC.
If the system's final temperature is 46oC, what was the initial temperature of the lead?
T = ? oC Pb
mass = 322 g
Ti = 25oC
mass = 264 g Pb
Tf = 46oC -
LOSE heat = GAIN heat
- [(Cp,Pb) (mass) (DT)] = (Cp,H2O) (mass) (DT)
Drop Units:
- [(0.138 J/goC) (322 g) (46oC - Ti)] = (4.184 J/goC) (264 g) (46oC- 25oC)]
- [(44.44) (46oC - Ti)] = (1104.6) (21oC)]
Ti =
44.44 Ti =
Ti = 568oC
Calorimetry Problems 2
question #12

50. of the system is 24oC, what was the mass of the ice?
A sample of ice at –12oC is placed into 68 g of water at 85oC. If the final temperature
of the system is 24oC, what was the mass of the ice?
T = -12oC
H2O
mass = ? g
Ti = 85oC
mass = 68 g
ice
Tf = 24oC
GAIN heat = - LOSE heat
qA = [(Cp,H2O) (mass) (DT)]
qA = [(2.077 J/goC) (mass) (12oC)]
24.9 m
[ qA + qB + qC ] = - [(Cp,H2O) (mass) (DT)]
[ qA + qB + qC ] = - [(4.184 J/goC) (68 g) (-61oC)]
qB = (Cf,H2O) (mass)
qB = (333 J/g) (mass)
333 m
458.2 m =
458.2
qC = [(Cp,H2O) (mass) (DT)]
qC = [(4.184 J/goC) (mass) (24oC)]
100.3 m
m = g
qTotal = qA + qB + qC
458.2 m
Calorimetry Problems 2
question #13

51. Endothermic Reaction +DH Endothermic Energy Reaction progress
Energy + Reactants  Products
Activation
Energy
Products
Energy
+DH Endothermic
Reactants
Reaction progress

52. Calorimetry Problems 1 Calorimetry 1 Calorimetry 1 Keys

53. Calorimetry Problems 2 Calorimetry 2 Calorimetry 2
Specific Heat Values
Calorimetry 2 
Specific Heat Values
Keys

54. Heat Energy Problems Heat Energy Problems Heat Energy Problems
  Heat Problems (key)         
Heat Energy of Water Problems (Calorimetry) Specific Heat Problems
Heat Energy Problems
  Heat Problems (key)         
Heat Energy of Water Problems (Calorimetry) Specific Heat Problems
Keys
a b c

55. Enthalpy Diagram H2(g) + ½ O2(g) H2O(g) H2O(l) Energy
DH = +242 kJ
Endothermic
-242 kJ
Exothermic
-286 kJ
Endothermic
DH = -286 kJ
Exothermic
H2O(g)
Energy
44 kJ
Exothermic
+44 kJ
Endothermic
H2O(l)
H2(g) + 1/2O2(g)  H2O(g) kJ DH = kJ
Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211

56. Hess’s Law
Calculate the enthalpy of formation of carbon dioxide from its elements.
C(g) + 2O(g)  CO2(g)
Use the following data:
2O(g)  O2(g) DH = kJ
C(s)  C(g) DH = kJ
CO2(g)  C(s) + O2(g) DH = kJ
2O(g)  O2(g) DH = kJ
C(g)  C(s) DH = kJ
C(s) + O2(g)  CO2(g) DH = kJ
C(g) + 2O(g)  CO2(g) DH = kJ
Smith, Smoot, Himes, pg 141

57. A team that gains 10 yards on a pass play but has a five-yard penalty,
In football, as in Hess's law, only the initial and final conditions matter.
A team that gains 10 yards on a pass play but has a five-yard penalty,
has the same net gain as the team that gained only 5 yards.
10 yard pass
pg 353 Holt Chemistry Myers, Oldham, Tocci , 2004 (Holt Rinehart)
5 yard net gain
5 yard penalty
initial position
of ball
final position
of ball